# How do I prove that $\det A_{n+2} = a \det A_{n+1} + b \det A_n$ for matrix $A$?

I have calculated:

$\det A_1=2$, $\det A_2=3$, $\det A_3=4$, so I was putting some numbers in $\det A_{n+2} = a \det A_{n+1} + b\det A_n$ like $n=1$, $n=2$ ($\det$ $n\times n$ matrix) and get that $a=2$, $b=-1$, but how that can help me to prove that $\det A_{n+2} = a\det A_{n+1} + b\det A_n$?

And the end how do I prove that $\det A_n = n+1$?

it's easy indeed. to calculate $\det(A_{n+2})$ expand determinant from first column and you get this: $$\det(A_{n+2})=2\det(A_{n+1})-\det(B)$$ where $\det(B)$ is $\det(A_{n+2})$ with the first column and second row removed. now again expand $\det(B)$ in first row to get $\det(B)=\det(A_{n})$ and proof is complete.
to solve the second part $\det(A_{n})=n+1$, can be done by induction.
Use induction on $n$. For $n=1$, the result is trivial. Now, suppose that there is an $N\in\mathbb{N}$ such that $\det A_n=n+1$ for all $n\le N$. Then for the case $N+1$, we have \begin{align} \det A_{N+1} &=2\det A_N-\det \begin{pmatrix} 2&1&0&0&&&&&\\ 1&2&1&0&&&&&\\ 0&1&2&1&&&&&\\ &&&&\ddots\\ &&&&&1&2&1&0\\ &&&&&0&1&2&0\\ &&&&&0&0&1&1 \end{pmatrix}_{N\times N}\\ &=2(N+1)-\det \begin{pmatrix} 2&1&0&0&&&\\ 1&2&1&0&&&\\ 0&1&2&1&&&\\ &&&&\ddots\\ &&&&&1&2&1\\ &&&&&0&1&2 \end{pmatrix}_{(N-1)\times(N-1)}\\ &\quad+\det \begin{pmatrix} 2&1&0&0&&&\\ 1&2&1&0&&&\\ 0&1&2&1&&&\\ &&&&\ddots\\ &&&&&1&2&0\\ &&&&&0&1&0 \end{pmatrix}_{(N-1)\times(N-1)}\\ &=2N+2-[(N-1)+1]+0\\ &=N+2. \end{align} This completes the proof for the second question.